$r_1 + r_2 = 75^\circ$ For $TIR$ at back surface $\sqrt{3} \sin r_2 = \frac{3}{2} \sin 90^\circ$ $r_2 \geq 60^\circ$ $r_1 \leq 15^\circ$ $1 \sin

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$r_1 + r_2 = 75^\circ$ For $TIR$ at back surface $\sqrt{3} \sin r_2 = \frac{3}{2} \sin 90^\circ$ $r_2 \geq 60^\circ$ $r_1 \leq 15^\circ$ $1 \sin i = \sqrt{3} \sin 15^\circ$ $\sin i = 1.73 \times 0.25$ $\sin i = 0.433$ $i = 25^\circ \Rightarrow i < 25^\circ$

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

A prism of angle $75^\circ$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index $1.5$ only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be . . . . . . .
$(\sin 15^\circ = 0.25$ and $\sin 25^\circ = 0.43)$

JEE MAIN 2026 All JEE MAIN

A
between $15^\circ$ and $20^\circ$
B
$15^\circ$
C
$> 25^\circ$
D
$< 15^\circ$

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A prism of angle $75^\circ$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index $1.5$ only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be . . . . . . . $(\sin 15^\circ = 0.25$ and $\sin 25^\circ = 0.43)$

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A prism of angle $75^\circ$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index $1.5$ only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be . . . . . . .
$(\sin 15^\circ = 0.25$ and $\sin 25^\circ = 0.43)$

Using the Trapezoidal rule,find the approximate value of $\int_1^4 y \, dx$ based on the following table:
$x$ $1$ $2$ $3$ $4$
$y$ $0.7111$ $0.7222$ $0.7333$ $0.7444$
(in $.1833$)